Answer:
B=![\left[\begin{array}{ccc}0&0\\0&1\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%260%5C%5C0%261%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
Let's do the multiplication AB.
If A=![\left[\begin{array}{ccc}1&0\\0&0\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%5C%5C0%260%5C%5C%5Cend%7Barray%7D%5Cright%5D)
then the first row of A is= (1 0) by the first column of B= (0 0) is equal to zero.
the first row of A is= (1 0) by the second column of B= (0 1) is equal to zero too because 1.0+0.1=0.
the second row of A is= (0 0) by any colum of B is equal to zero too.
So we have found an example that works!
Answer:
The answer is 113.04
Step-by-step explanation:
We cannot use CPCTC to show that BE is congruent to DE.
CPCTC refers to the angles and sides of two congruent triangles being congruent to their corresponding pieces. The height of a triangle is not one of the given sides that counts in this.
Answer:
Solution: x = 6
Step-by-step explanation:
Given equation is:

In order to solve the equation both sides will be squared

Verifying the solution
